Question

Find the dimensions of ${\epsilon }_0{E}^2$ (${\epsilon }_0$=permittivity in vacuum , E = electric field).

• A. $M^1{L}^{-1}{T}^{-2}$
• B. $M^1{L}^1{T}^{-2}$
• C. $M^{-1}{L}^{-1}{T}^{-2}$
• D. $M^1{L}^{-1}{T}^{-1}$

Answer 1

The correct option is A $M^1{L}^{-1}{T}^{-2}$

Energy density = $\frac{1}{2}{\epsilon }_0{E}^2$

[Energy density] = $\left[{\epsilon }_0{E}^2\right]$

$\left[\frac{1}{2}{\epsilon }_0{E}^2\right]=\frac{\left[energy\right]}{\left[volume\right]}=\frac{M^1{L}^2{T}^{-2}}{L^3}=M^1{L}^{-1}{T}^{-2}$